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August 4

Bond Energy

Hello. Under which conditions (STP or SATP) is the bond energy of a single carbon-to-carbon bond 347 kJ mol^-1? Thanks in advance. --Mayfare (talk) 00:54, 4 August 2009 (UTC)[reply]

As it's chemistry/physics it will use the IUPAC conditions http://goldbook.iupac.org/goldbook/S06036.html

If you want to KNOW FOR CERTAIN then you need to find and read the reference from which the data came.

Also are STP and SATP actually different?83.100.250.79 (talk) 14:23, 4 August 2009 (UTC)[reply]

STP is 0°C and 101.325 kPa. SATP is 25°C and 100 kPa. --Mayfare (talk) 18:36, 4 August 2009 (UTC)[reply]

Isn't it the other way round (1atm-101.325kPa) if A means 'atmospheric'.

I'm not sure that the acronym STP has a single defined value - it may depend on whether you are a chemist, physicist or other type of scientist.83.100.250.79 (talk) 20:16, 4 August 2009 (UTC)[reply]

STP stands for Standard Temperature and Pressure. SATP stands for Standard Ambient Temperature and Pressure. Sorry for being unclear in the first place. --Mayfare (talk) 18:35, 5 August 2009 (UTC)[reply]

I read this article on standard conditions for temperature and pressure. I did not realize that I was using US standards. --Mayfare (talk) 17:50, 6 August 2009 (UTC)[reply]

Peter Pan Syndrome - refusing to heed written laws or just societal norms?

Reading the article in our link on the above, I was curious. When People refer to this, are they talking about people who refuse any type of restrictions, especially legal ones? Or, are they talking mostly cultural ones, not wanting to take the responsibilities that adults generally do.

What brought this up was a monarch - Ludwig II of Bavaria whom I had heard elsewhere might been like this (I hesitate to use 'suffered" since it's not an actual diagnosis), because of the fantasy world in which he often seemed to live. Although, I imagine just "living in a fantasy world" isn't the only criteria, it certainly seems to be part of it. Or, does the term generally refer to someone a little more dangerous or odious than just someone "living in a fantasy world"?209.244.30.221 (talk) 01:08, 4 August 2009 (UTC)[reply]

In a social setting (i. e. not a medical one), I have only ever heard this term used to describe men who never seemed to be interested in taking on adult responsibility. (Please note that I am not suggesting that this only happens to males, but simply that "Peter Pan" is usually applied only to males.) I have never heard the use to mean anything illegal or odious, unless the idea of perpetual childhood is an odious one. Usually it is women making the complaint, and it is a complaint. // BL \\ (talk) 01:36, 4 August 2009 (UTC)[reply]

As our article says - this is a "pop-psych" term - it's not a proper medical term with a hard definition. So what do people mean when they use it? Well, it's just some vague concept that an adult behaves like a kid. Different people are bound to use the word in different ways. SteveBaker (talk) 03:30, 4 August 2009 (UTC)[reply]

Old refrigerators making more ice

I have an old refrigerator and it makes more ice as when it was new. Now it needs to be defrosted at regular intervals. Why? Wouldn't it be much more logical if it made less ice as it gets older?--80.58.205.37 (talk) 11:01, 4 August 2009 (UTC)[reply]

My guess is that it's because with age the seal becomes less effective, so more water vapour gets inside the fridge. AndrewWTaylor (talk) 11:05, 4 August 2009 (UTC)[reply]

I agree, maybe the gasket in the door is old, so outside air leaks in and carries moisture with it. It could also be that different stuff give off different amonunts of moisture (if your fridge held two sixpacks of beer a few years ago and today it holds vegetables, home made baby food and leftovers you will probably see some difference in the amounts of ice). Also if you open the door more often you let more air in.Sjö (talk) 11:10, 4 August 2009 (UTC)[reply]

Yep - your refrigerator is still getting cold enough to freeze the moisture out of the air. Usually, what happens is that the air inside the fridge is first cooled to the point where it can't hold so much water. That causes the water to condense onto the cold surfaces - just like the windows on your car fogging up when it's cold outside. The water droplets that end up on the freezer compartment then freeze into ice. However, if all was well, there would now be no more water in the air inside the fridge and no more ice would form. But if the door seal leaks (especially if it's towards the bottom of the door) then the cold/dry air (being denser) will slowly flow out of the fridge to be replaced by moist/warm air from outside. The warm air carries with it more moisture into the fridge and the cycle repeats, gradually building up the ice. Replacing the door seal should stop that and will save you money too. Failing that (as others have suggested) it would have to mean some change in life-style...perhaps different foods being stored, open containers of liquids - increased opening of the door - or perhaps a new family member who doesn't shut the fridge door firmly enough or soon enough. I suppose it's also possible that you have the thing dialled down to a colder setting than before, or that the thermostat has gone wonky...but that doesn't explain where the water to form the ice is coming from...so I'm still betting that the door seal is failing to keep the thing airtight. You should consider changing the seal though - it's a real waste of electricity. But at least you know that your old fridge is still capable of keeping things properly cool! SteveBaker (talk) 12:06, 4 August 2009 (UTC)[reply]

To check the door seal, use a piece of ribbon. Close the door on it and see if it can be pulled out freely. Check several places around the door, including the hinged side. —Preceding unsigned comment added by 98.21.108.4 (talk) 13:48, 5 August 2009 (UTC)[reply]

P.S. Instead of a ribbon, a narrow strip of paper would be better. This has the stiffness to allow it to be inserted easily in the sides of the door. —Preceding unsigned comment added by 98.21.106.176 (talk) 13:22, 6 August 2009 (UTC)[reply]

No - you don't insert the paper into the closed door - you open the door, place the ribbon/paper against the seal, then close the door and see how hard it is to pull it back out again. If there is a significant gap in the seal, the ribbon/paper will come out easily. It's gonna be hard to find a problem that way though. I'd get one of those electronic temperature probes and just look for cold spots on the outside of the seal when the door has been closed for an hour or two. Personally - I wouldn't bother - if the problem is all that serious, it's going to be a defective seal for 100% sure...so just replace it already! SteveBaker (talk) 13:46, 6 August 2009 (UTC)[reply]

does the acidity of aromatic rings, alkenes (or even alkanes) contribute to petroleum formation?

So, the pKa of an alkane is like 60, and that of alkene around 45, .... but that means a proton still comes off occasionally. That starts to make me think ... if I say stored a pure olefin (or maybe benzene) in a glass jar for 1000 years, would I eventually see some signs of chemical reactions (maybe on the ppm scale ... if someone was alive to run the jar through NMR later?). I'm thinking carbon-carbon bonds would be formed in this way ... well, let's say we had liquified 1-butene or something, ever so often, the boltzmann distribution apparently gives a lucky butene molecule enough energy to lose a proton ... a proton which then proceeds to readily protonate some other butene .... which makes it a cation, which then finds the anion, forming a C-C bond. (Well, it doesn't have to find the original molecule that lost the proton, the original anion would probably have pulled a proton off some other molecule later on, creating a new anion....)

Yeah, it probably would occur very very slowly ... but then it occurs to me, that would probably happen in an oil bed under high heat and pressure, given enough time (like 50 million years). I'm using a pure olefin as a purely hypothetical thought experiment of course (to allow ease of detection of new products), since I assume in peat or whatever you have a wide collection of organic olefins. Or is the breakage and reformation of C-C bonds (homolytically or hemolytically) under high heat and pressure the more predominant process? John Riemann Soong (talk) 11:42, 4 August 2009 (UTC)[reply]

I don't know the answer, but Petroleum#Formation and Catagenesis (geology) say that C-C bonds are formed by living organisms, which die and become kerogen. Kerogen subsequently undergoes thermal decomposition to hydrocarbons.

No mention of acid-base reactions. The pK_a values of alkanes and alkenes are very approximate and in any case depend on the solvent they're in.

Ben (talk) 12:07, 4 August 2009 (UTC)[reply]

The loss of a proton, or H radical could contribute to petroleum formation, as will all conceivable chemical processes including dehydration, other eliminations, pericyclic additions rearrangements and eliminations etc.83.100.250.79 (talk) 14:28, 4 August 2009 (UTC)[reply]

They could in principle, but do they in practice?

The answer may not be known.

Ben (talk) 14:34, 4 August 2009 (UTC)[reply]

By the way peat etc is predominately polysaccharides/cellulose (and lignin) and one of the major reactions to get hydrocarbon is loss of Oxygen (possible as water), additionally the carbon skeleton of 'peat' is not the same as oil/coal so C-C bond changes must be a major factor, along with C-H and C-OH bond changes. (It's heterolytic cleavage not hemolytic)83.100.250.79 (talk) 16:41, 4 August 2009 (UTC)[reply]

Have a look at this lignin structure - and compare that with coal (a fossil fuel)..83.100.250.79 (talk) 16:44, 4 August 2009 (UTC)[reply]

Wow that lignin seems to have some ring strain ... I mean, I like how two substituents on a benzene ring can later "join up" later on.... and just how does it react that it fuses multiple aromatic rings together? That's amazing. John Riemann Soong (talk) 18:15, 4 August 2009 (UTC)[reply]

May I add that peat and lignite also form biologically, from plant matter? 98.234.126.251 (talk) 01:39, 5 August 2009 (UTC)[reply]

The contrast between the two structures is interesting... looking at the disulfide and nitrogen linkages in the coal structure, it's pretty clear that protein is a major precursor, while lignin is essentially purely carbohydrate, which is what you would expect from collulose-like matter. Hmm... I wonder how common these patterns are among the compounds. – Clockwork Soul 01:55, 5 August 2009 (UTC)[reply]

ok Lignin is not pure carbohydrate at all (it's not a cellulose)- carbohydrate has not bezene rings/polyphenols (there are good structures of carbohydrates at cellulose)

As for the N's and S's in coal - I would guess that they do derive from proteins - but possibly by decomposition (eg anaerobic bacteria action to give H2S, or NH3), though it could equally be by direct interaction of proteins and other plant matter. This article suggest that both methods may contribute [1] - (though don't take that as fact)

There's a mention of the origins and types of sulphur in coal in this thesis http://witsetd.wits.ac.za:8080/dspace/bitstream/123456789/7060/4/EL%20Koper%20PhD%20(c)%202009%20-%2003%20Background%20and%20reviews.pdf (which I can't find a title for but seems to have been writen by a EL Koper) There probably are better descriptions on the web if you look, maybe in google books.83.100.250.79 (talk) 13:05, 5 August 2009 (UTC)[reply]

I'm no sure the disulfide linkages in coal mean anything about it coming from "protein". Given the sort of chemical changes that occur between the living tissue and the fossil fuels, there really should not be any recognizable biomolecules there, and any "evidence" of such biomolecules should merely be the chemical equivalent of "convergent evolution". If there is sulfur and carbon present in the mix, then disulfide linkages are likely to form regardless of the source of the sulfur and carbon. I would say that it is impossible to consider that the disulfide linkages found in protein (via cystein) would remain unchanged through the harsh conditions that created coal from decayed biological material. --Jayron32 02:56, 6 August 2009 (UTC)[reply]

I do see an amazing resemblance between chlorophyll and whatever residue was found (mentioned in the petroleum formation article). John Riemann Soong (talk) 03:21, 6 August 2009 (UTC)[reply]

"ok Lignin is not pure carbohydrate at all (it's not a cellulose)- carbohydrate has not bezene rings/polyphenols (there are good structures of carbohydrates at cellulose)" -- Benzene rings, polyphenols and other aromatic-type compounds in coal (lignite, whatever have you) form by thermal dehydration / dehydrogenation of carbohydrate units in lignin and cellulose (that's pretty obvious, when you think about it). FWiW 98.234.126.251 (talk) 06:02, 6 August 2009 (UTC)[reply]

Lignin is a compound in wood - it's formed in the living tree, and already contains the phenols whilst it is still alive. (see that article)83.100.250.79 (talk) 14:20, 6 August 2009 (UTC)[reply]

Yeah, I know that, what I'm saying is that cellulose is also subject to thermal dehydration/dehydrogenation to form polyphenolic and polyaromatic compounds. (In any case, wood contains more cellulose than lignin.) 98.234.126.251 (talk) 23:04, 6 August 2009 (UTC)[reply]

Interstellar travel

In the far future, when travel to distant stars is possible or even common, will it be easier/faster to travel along the comparatively crowded arms of the galaxy or along the emptier spaces in between?

88.108.8.64 (talk) 13:15, 4 August 2009 (UTC)[reply]

Maintaining a constant speed while travelling from one location to another requires no energy if no other forces act upon a body, so it should be equally easy for both (though travelleing intergalactically would take considerably longer). However, the common theme in sci-fi is that to travel faster than light you need to maintain something (a warp field, a subspace bubble, etc) which is a constant drain on energy. So going somewhere closer would be easier. I'm not sure exactly how crowded the arms of our galaxy are, but if I recall correctly, you could travel in a straight line to a nearby star without expecting to hit anything along the way.

But all of that is kind of irrelevant: since we don't know how to travel faster than light, we can't say how difficult it is. Vimescarrot (talk) 13:22, 4 August 2009 (UTC)[reply]

Assuming real-world (non warp magic) physics, which way you'd go depends greatly on the technology of your spaceship and its drive system. The advantage of the space between arms is that there's a bit less debris - once you're travelling at relativistic velocities then impacting even a modest sized particle can be damaging, forcing you to have quite a robustly constructed spaceship (this assuming that there really is less matter between the arms than in them). The advantage of being in the arm is that if you rely on there being free matter (e.g. if your ship is propelled by something like a Bussard ramjet) then you'll have more to chew on in the arms. But really we're so far from being able to to anything like this, that the engineering details are really anybody's guess (it'd be like arguing with Da Vinci about whether his hang-glider thing would be better than his helicopter thing; you don't know until you build one). -- Finlay McWalter • Talk 13:48, 4 August 2009 (UTC)[reply]

How much variation is there in density between the arms and the 'gaps' (in the plane, at a given distance from the core)? I've gotten the impression that the bright 'arms' are merely where a pressure wave passed recently, creating young bright stars. —Tamfang (talk) 02:09, 11 August 2009 (UTC)[reply]

I was thinking more in terms of the effect of gravity, similar to how current space probes and such like are sent close to certain planets on their way somewhere else, helping them go faster. Though I suppose not crashing into things would also be useful. 88.108.8.64 (talk) 14:09, 4 August 2009 (UTC)[reply]

The stars are probably too far apart for you to gravity assist in a useful fashion. A gravity assist will not get you faster then light, and if you have to travel 5 years to get to the star to do the gravity assist that will increase your speed by 10%, is that really worth it? Googlemeister (talk) 14:23, 4 August 2009 (UTC)[reply]

In that case you might like to read gravity assist and Interplanetary Transport Network; but those only describe intra-solar system travel, and even then taking a decade to get anywhere. It'd take millennia to move between stars by this mechanism. -- Finlay McWalter • Talk 14:24, 4 August 2009 (UTC)[reply]

I just want to point out, since it is assumed but not addressed directly in the comments above, that most people here have been answering in a way that assumes faster-than-light is probably the only way to do this efficiently. There are slower-than-light approaches but they require a ship that takes decades and decades if not centuries to get from point A to point B (which, even if it really is possible for humans to do that—which I'm not convinced—cannot certainly be a "common" activity). This is because the vastness of the universe is, well, VAST. The distances are HUGE. If we cannot travel faster-than-light, there is little likelihood of any kind of Star Trek future for us. Even traveling at the speed of light is pretty slow on intergalactic scales, compared to the span of human lives (or, worse, the span of human attentions). --98.217.14.211 (talk) 15:10, 4 August 2009 (UTC)[reply]

Relativity can (mostly) fix the timespan issue for the passengers, though. If you posit a spaceship that can withstand debris at relativistic speeds (a Bussard ramjet above is a good starting point), then you don't need a multi-generational ship even if Earthbound observers note a multi-generational journey. This still prohibits a Star Trek-type future, but not a humans-in-space future. Poul Anderson's Starfarers is a good treatment of the subject. — Lomn 15:25, 4 August 2009 (UTC)[reply]

It depends how far you are going to travel, obviously, and whether you want to have any kind of communication with Earth (which is basically prohibited in any useful way). Even with relativity, I find it unlikely that you could get many humans to sign up for 20 years on a ship. It's not Star Trek; you're going to read all the books you have pretty quickly, have all the conversations you can have pretty quickly, and the stars are going to get a bit dull. Even a five year trip would be quite disruptive in the course of one's life, if nothing was happening on it (it would be one thing if you were traveling around the world, a different port every night... but in space it's a lot more monotonous). (As you may be able to tell, I am quite pessimistic about space travel without the possibility of FTL. I tend to think that those who push most strongly for it are just being escapist.) --98.217.14.211 (talk) 18:00, 4 August 2009 (UTC)[reply]

So no chance then of any particular route across the galaxy being much better for long distance transit? I was hoping for some sort of trans-galactic path with lots of starships running back and forth through it.88.108.8.64 (talk) 17:16, 4 August 2009 (UTC)[reply]

Not without faster than light technology. If you have that, well, that changes things, but since we don't know what that truly would look like (since there is not the slightest indication that it is possible), it's hard to say. --98.217.14.211 (talk) 18:00, 4 August 2009 (UTC)[reply]

Certain routes may well turn out to be better than others, but I can't see lots of starships covering interstellar distances without faster-than-light travel. You don't start journeys that are going to last years very frequently. I would be surprised if craft passed other craft more than a handful of times during their journey. --Tango (talk) 20:57, 4 August 2009 (UTC)[reply]

I've talked about this many times before - so I'll keep the explanation short. There IS a way...at least in theory. You have to transfer your brain into a computer...build a computer that can replicate in great detail the precise functioning of every neuron every chemical pathway. This concept doesn't violate any fundamental laws - and many people believe it will be possible in the not too distant future. You arrange that your entire psyche - everything that makes you be "you" is in the machine - and then you destroy your physical body. OK - now you can put your brain into a computer on board a very slow spaceship - and adjust the clock rate of the computer such that the computer program that is your brain runs very slowly (You could install Windows Vista, for example!)...you will be "thinking" very slowly. For you, inside the computer, time can now be speeded up and (to some degree) slowed down at will. So - off you go on your million year (thousand lightyear) trip - and it seems to you like it only took half an hour maybe...if you see something interesting along the way - you can temporarily speed up the clock on your computer...spend as long as you want observing whatever it is...and then slow the clock down again to 'fast forwards' over the boring parts. So long as we can make spacecraft that are reliable enough - speed is a relatively insignificant barrier for "humans" (gotta use the quotes to keep everyone happy!) to colonise the entire galaxy. When you get where you're going, you put your brain/computer into a realistic humaniod robot and you can carry on with life more or less as usual. Of course, for very long trips, you have the problem that the place you saw through the telescope (which is already a very out of date view) will have changed considerably by the time you get there. But it's not an impossible prospect. If the spacecraft you're using is unreliable - you can send out a spacecraft with an 'empty' computer. When it gets where it's going, it can send you a speed-of-light radio message to say it's OK - then you can send your brain software as a digital data stream back to the computer. In effect, you can travel at the speed of light...so long as there is a suitable computer at the other end. But for you, it would seem just like instantaneous teleportation. Your robotic self climbs into the booth - you dial up some far distant star that humanoid robots have already been to - you push the button and in literally zero time (for YOU), you can step out in an identical robotic body at the other end. Of course the actual elapsed time would be vast...but maybe you don't care about that. SteveBaker (talk) 22:26, 4 August 2009 (UTC)[reply]

Rather than putting your consciousness in a computer, you could put your body in stasis for the duration of the journey. Quantumelfmage (talk) 21:15, 10 August 2009 (UTC)[reply]

You don't really need any special computer technology to do this, as Lomn pointed out. Travelling in a ship at very near light speed, time goes by much slower for the passengers than for an observer on Earth. For example if you had a ship that could accelerate to 0.9999995c relative to its initial frame in a relatively short time period, you could travel 1000 light years from Earth and only experience 1 year passing. Of course if you made the return trip, everyone you knew on Earth would be long dead. Rckrone (talk) 00:45, 5 August 2009 (UTC)[reply]

While there are no theoretical problems with near-light speed travel, there are several engineering problems that we are nowhere near solving (some means of propulsion and a way of surviving hitting interstellar dust particles at such high speeds being the main ones). Steve's idea may be more achievable. We have ever improving computer systems, it probably won't be long before they are powerful enough to simulate the human brain, we just need a better understanding of how the human brain works. Personally, I think we should just be content with travelling to nearby stars, where less exotic means of transport would suffice (travelling at 0.1c there are several stars within a lifetime's travel of us and at those speeds relativistic effects aren't too serious - you need to account for them in your calculations, but you don't need to worry too much). --Tango (talk) 01:07, 5 August 2009 (UTC)[reply]

But, but...but...I want my instantaneous (to me) interstellar transporter! I want to teleport to some planet orbitting Proxima Centauri - spend a few days looking at the sights - then teleport back again. Sadly, I'll need 8 years off work in order to do it...but it would certainly be worth the trip! SteveBaker (talk) 01:20, 5 August 2009 (UTC)[reply]

You're a computer graphics expert - make a virtual Proxima Centauri. What is the difference between a real you observing a virtual star and a virtual you observing a real star? They seem equally good to me, except the former is far easier and doesn't require 8 years off work. --Tango (talk) 02:02, 5 August 2009 (UTC)[reply]

Do you take the red pill or the blue pill? SteveBaker (talk) 15:05, 5 August 2009 (UTC)[reply]

What's this about humans not wanting to sign up for 20 years on a ship? Guys, people in the medieval Age of Sail, or on the Silk Road had it far worse. Entire nomadic groups would take decades to move to one place to another. What's worse? Being imprisoned for 20 years on a galley. John Riemann Soong (talk) 07:55, 5 August 2009 (UTC)[reply]

I don't think a nomadic group on the Silk Road is a very good comparison to interstellar travel. You could walk the entire silk road in about 1 year, and while it isn't a stroll through Paris, there is a whole lot more to see on the way then on a 20 year trip through space with nothing in it. Googlemeister (talk) 12:56, 5 August 2009 (UTC)[reply]

Plus, with any kind of realistic technology, you can't even get out of the solar system in 20 years. People who signed up for 20 years on board ship would get shore leave every six months or so. The guys on the spaceship can't even look out of the window and see anything interesting. SteveBaker (talk) 15:05, 5 August 2009 (UTC)[reply]

Plenty of realistic technologies can get out of the solar system in under 20 years. They don't exist yet, but they are perfectly realistic. Laser beam propulsion is probably the most likely to happen. I wouldn't be surprised if an unmanned flyby probe is sent to the Alpha Centauri system within my lifetime (it might not get there within my lifetime, though...). --Tango (talk) 20:45, 5 August 2009 (UTC)[reply]

I sure wouldn't want to sign up on a laser-beam-propelled spaceship (or indeed on any spaceship without a self-contained propulsion system) -- there's always a serious risk that by the time you get to where you're going, the laser-beam apparatus would have been de-funded for financial / political reasons and you got no way to get back home again. 98.234.126.251 (talk) 02:24, 6 August 2009 (UTC)[reply]

I think manned interstellar travel, at least at first, would be a one way trip anyway. While you can get to nearby stars in a lifetime at, say, 0.1c, you can't get back again (and if it is a generation ship you would probably be planning to establish a colony which could build its own beam emitter for any return journeys). More of an issue with beamed propulsion is deceleration at the other end. You can do that using a mirror to reflect the beam back at the front of your craft, but only if they turn the beam on at the right time. You could decelerate a little using a conventional solar sail (almost like parachuting into the new star system), but probably not from near-relativistic speeds. That's why the probe I think might be launched in my lifetime would be a flyby, it would be too difficult to slow it down so it could be captured. --Tango (talk) 02:51, 6 August 2009 (UTC)[reply]

The Migration Period lasted for centuries, and I don't mean to imply that permanent nomadism does not exist, but I doubt that anyone ever planned a definite route to take more than a few years. "We'll cross the Bering Strait, then our grandchildren will reach Vancouver Island, and their great-grandchildren will drink margaritas in Cancún" – nope. —Tamfang (talk) 02:09, 11 August 2009 (UTC)[reply]

This generation might not like to sign up .... but lifestyles can change, and culture can adapt. Would you really want to be a pioneer and own a log cabin out in the middle of nowhere, where you might not see town for another 30 years? Perhaps occasionally (as you're being flung forwards), you could get out of the ship and stretch (well, as long as you somehow maintained velocity with the ship). John Riemann Soong (talk) 02:30, 6 August 2009 (UTC)[reply]

If I were planning a long interstellar voyage, especially if it were a generation ship, I would send about 3 ships at once. It would allow those on one ship to evacuate to the others in an emergency (perhaps not permanently, but for a couple of days while life support was repaired - what you do if it turns out to be irreparable, I'm not sure...). It would also help reduce the feeling of isolation since you wouldn't be alone in the interstellar void, you would be part of a formation and could travel between the ships. --Tango (talk) 02:51, 6 August 2009 (UTC)[reply]

You mean, sort of like Columbus and his 3 caravels? 98.234.126.251 (talk) 03:18, 6 August 2009 (UTC)[reply]

I've no idea. Not being American, I've never learnt much about Columbus. --Tango (talk) 04:00, 6 August 2009 (UTC)[reply]

I'm curious -- can you perform gravity assists (not necessarily stars -- perhaps large Kupier Belt objects, etc.) for minor trajectory corrections for a fleet of ships of different masses while keeping them altogether? I know at a certain distance away from the object, their gravitational accelerations are all going to be the same, but their kinetic energies won't be. John Riemann Soong (talk) 05:21, 6 August 2009 (UTC)[reply]

If their accelerations are the same, then they'll stay together. FWiW 98.234.126.251 (talk) 05:54, 6 August 2009 (UTC)[reply]

"Would you really want to be a pioneer and own a log cabin out in the middle of nowhere, where you might not see town for another 30 years? Perhaps occasionally (as you're being flung forwards), you could get out of the ship and stretch (well, as long as you somehow maintained velocity with the ship)." -- That's not the issue that I was talking about. The problem specific to a laser-beam propelled ship is that it's dependent on an external laser installation inside the Solar System to propel it to its destination and back again (similar to, say, a cable car being dependent on a lineside winch to propel it forward). Now, it's a given that the journey will take many, many years, during which time the laser installation will have to be maintained by us Earthlings at considerable expense so it could generate the laser beam again to propel the ship back to the Solar System on its return journey. Well, during that time it's very likely that some cost-cutting politician will come along and say, "Why are we spending all that money maintaining that laser installation that ain't doing nothing? Let's cut it from the budget and save some cash" -- and if that happens, the astronauts on that spaceship will be stranded in some galaxy far far away with no way to get back home. As for "this generation might not like to sign up", etc., etc. -- well, for one thing I might sign up if they paid me enough for it, but ONLY IF THE SHIP GOT ITS OWN PROPULSION ENGINE ON BOARD!!! 98.234.126.251 (talk) 03:03, 6 August 2009 (UTC)[reply]

I don't think John was replying to you. A reply is usually indented one tab more than what it is replying to, not indenting at all usually means it is a general reply to everything that has gone above or a completely new point. I've moved your reply to the bottom and undone the change you made to the indentation of my reply - people don't alter other people's comments and keep comments at the same level in chronological order. --Tango (talk) 04:00, 6 August 2009 (UTC)[reply]

I thought he was. My mistake. (BTW, I added a quote from John's post that I was replying to, so's it's clear who's replying to whom.) 98.234.126.251 (talk) 04:31, 6 August 2009 (UTC)[reply]

For a ship going at a substantial fraction of 'c', the time dilation effects would mean that more time was passing on earth than on-board the ship. But even without that - over a period of decades, there are wars, natural disasters, all kinds of things that could stop that laser from working. But in any case, to be practical, the laser would have to be set up in it's own solar orbit with gigantic solar panels providing the power source for it. We're not talking about a small laser here. To transmit enough power to push a spacecraft along at a reasonable acceleration, we're talking about a laser that would look MUCH brighter than the sun at the distance of the earth's orbit with a beam diameter of a couple of kilometers to cover the large solar sail that the spacecraft would need in order to be able to keep it cool enough. Out in interstellar space, this thing is by far the brightest thing in the sky even when you're lightyears away from it. The laser would have to be pretty reliable because going out to fix it would be costly - and focussing it would require lenses out at the orbit of Mars or so because even lasers diverge by a tiny amount - and over lightyears, that dilutes it's power to a massive extent. Aiming the thing would be an interesting trick too...if the spacecraft ever somehow drifted out of the beam, it might have an impossible task finding it again and given the years it would take a radio message to return to earth to ask what happened to the laser...the mission would be in DEEP touble if that ever happened. SteveBaker (talk) 13:43, 6 August 2009 (UTC)[reply]

Indeed, using a laser here to decelerate a craft to be captured by another star or to accelerate it back towards Earth would be very difficult. For a flyby it's relatively easy, though, there is no need to keep beaming when it is lightyears away. Just propelling it for the first couple of years would probably be enough. Some back of the edit form calculations: To accelerate a 1 tonne probe to 0.1c requires (ignoring relativity) 4.5x10¹⁷J. If we do that over 2 years that requires a power of 7GW. At Earth's distance from the Sun (I would put the laser at Sun-Earth L4 or L5) solar power generates 1.4kW/m². Therefore we need solar panels 5 km² in area. That's big, but it's not impossible. --Tango (talk) 22:29, 6 August 2009 (UTC)[reply]

That seems to be assuming that the energy of an absorbed or reflected photon all turns into kinetic energy of the probe. I don't think you get within orders of magnitude of that. If you reflect the photon, most of its energy is carried away in the reflection; if you absorb it, it mostly just heats up the probe. Try doing it with momentum instead of energy. --Trovatore (talk) 22:40, 6 August 2009 (UTC)[reply]

The scarce resource is energy. Momentum is just a technicality based on the details of the design, which I don't have. It is also a more difficult calculation - the momentum transfer is going to vary over time as the relative velocities change, with energy you can just compare the initial state and the final state. I was assuming 100% efficiency, which is obviously unrealistic, but I was hoping it would get the right order of magnitude. I'll go and look it up... --Tango (talk) 23:37, 6 August 2009 (UTC)[reply]

You could use more of the energy if, say, you used it to power an ion drive, accelerating a propellant backwards at high velocity. The fraction of it you're going to get when you just bounce the photon backwards is tiny. 100% is not just unrealistic, it's wrong by orders of magnitude.

You might get a bit more of the energy once the craft is already moving at relativistic speeds. Then the photon is redshifted when you reflect it. Maybe you get to keep some of the difference between the incoming and outgoing photon; not really sure about that. This is a billiard-ball physics question, but it has to be done relativistically, which obviously makes it a bit more difficult. --Trovatore (talk) 23:42, 6 August 2009 (UTC)[reply]

Using the beam purely for power is an option, but it means you have to carry whatever you are going to use as exhaust. Even with the kind of exhaust velocities you get with an ion drive that's still a massive inefficiency on the scale we're talking about. If the photon being redshifted helped you could just start with a longer wavelength laser. I'm talking about speeds of around 0.1c, relativity isn't a big concern. If we only want orders of magnitude, Newtonian physics will suffice. --Tango (talk) 23:58, 6 August 2009 (UTC)[reply]

I sense that you're still missing the point, which is a pretty simple point and you're going to groan once you see it. The energy analysis that you've done is completely bogus. When you have a mirror sitting in space, and a photon of energy E hits it, does the mirror's kinetic energy increase by anything remotely close to E. No. The reflected photon has energy almost exactly E, and that's where almost all the energy goes.

Alternatively, you can absorb the photon, but in that case you get even less impulse, which should be obvious -- the photon's incoming momentum was E/c; if you reflect it, the impulse you get is the difference between that and the outgoing momentum −E/c, namely 2E/c. Whereas if you just eat the photon, the impulse you get is only E/c. So absorbing the photon is only half as good as reflecting it. --Trovatore (talk) 00:33, 7 August 2009 (UTC)[reply]

Yeah, I got that point after your last message. I've been trying to come up with a loophole to get me out of it. It all comes down to wavelengths - the reflected photon will be slightly redshifted and that is where the energy comes from. What determines how much it gets redshifted by? If you could redshift it enough you could get a large proportion of the energy. As we said, it is collisions of billiard balls. If we assume perfect reflection with no absorption then it is a perfectly elastic collision. If I can remember my A-level mechanics lessons [snip]. Ok. I can't remember by A-level mechanics lessons! I've just spent ages on it and can't get a useful answer. It seems, though, that you get a greater proportion of the energy if you use low energy photons, but I don't think the limit is 100% efficiency and I can't work out what it actually is (it seems to be dependant on the mass of the spacecraft, although I'm not sure in what way). Could someone else have a go at the maths? Also, can someone explain how the hell I just got a First Class MMath degree when I apparently can't do maths? Thanks! --Tango (talk) 03:12, 7 August 2009 (UTC)[reply]

Let's see. Using momentum instead of energy, 2E/c=mv, and plugging in the numbers gives 4.5*10^18 J: exactly 10 times the amount that Tango calculated. As for the redshift, a ship moving at 0.1c sees the laser photons coming in at approx. E(1-v/c); it reflects the photons, and Earth sees the reflected photons at E(1-v/c)^2. For v=0.1c, and considering the other uncertainties involved, we can simply forget about the redshifts and treat the photons as if they were reflecting off a stationary mirror. --Bowlhover (talk) 22:31, 7 August 2009 (UTC)[reply]

No, we can't. You can't assume the thing you are propelling is stationary, that's never going to work! There must be a flaw in your calculation somewhere - since the craft starts off stationary that would mean it gets zero energy per photon, it would never go anywhere. --Tango (talk) 00:22, 8 August 2009 (UTC)[reply]

The redshift can't be ignored for accurate computation, not even if you want a second significant figure, but for order-of-magnitude analysis (using momentum rather than energy) it can.

It's true to first order that the kinetic energy is not increasing at the start of the trip. K.E. is (1/2)mv^2, so the first derivative is mv(dv/dt), that is mva, where a is the acceleration. The acceleration actually diminishes as the craft speeds up, because it's reflecting redshifted photons which don't carry as much momentum -- but the rate of increase of K.E. goes up. --Trovatore (talk) 00:30, 8 August 2009 (UTC)[reply]

True, but I wasn't trying to determine the acceleration, I was trying to determine the efficiency. That is, of the energy that goes into sending the laser how much is used to accelerate the craft. That is all down to redshift. --Tango (talk) 17:24, 8 August 2009 (UTC)[reply]

It's interesting that Tango comes up with the number 4.5x10¹⁷J...that's almost exactly the number some OP quoted in a thread below for the energy produced by a decent-sized nuclear weapon. A 100 Megaton nuke (appropriately applied) would theoretically provide 4.2x10¹⁷J - enough energy to put that 1 ton capsule into an 0.1 c trajectory. 50 Megaton hydrogen bombs are certainly buildable with 1960's technology...perhaps the Project Orion approach is the right one after all? If we're talking about "Brain in a computer" transportation systems - or robotic probes of a more conventional kind...maybe that's a better way. SteveBaker (talk) 01:29, 7 August 2009 (UTC)[reply]

I heard they actually had a project for a nuclear-powered starship, called Project Daedalus. But the part that I just don't get is, how do they keep the nuclear explosions from melting the starship? 98.234.126.251 (talk) 02:18, 7 August 2009 (UTC)[reply]

Nuclear power plants don't melt (unless something goes wrong), so I it can it done. --Tango (talk) 03:12, 7 August 2009 (UTC)[reply]

It's a possible idea. You do, however, want to accelerate the ship without blowing it up. You would need lots of small nukes, not one big one. --Tango (talk) 03:12, 7 August 2009 (UTC)[reply]

Wait - everyone is getting confused. Project Orion was the seemingly crazy idea to build a spacecraft with a gigantic, heavy "pusher plate" at the back, then a bunch of huge shock absorbers and springs - and the payload/crew-quarters at the front. The idea was to toss small nuclear bombs out of the back of the machine and to explode them against the pusher plate. Project Daedalus was to use a continuous controlled fusion reaction - akin to the ideas use in experimental fusion power generators. Daedalus is technologically exceedingly difficult - we know that controlling and confining fusion reactions is really difficult. Orion seems absolutely crazy...totally insane...but when you crunch the numbers, it's quite do-able, even with present day technology. Well...do-able, IF you had the launch facility to get all of the thousands of tons of material needed for the pusher plate and shock absorbers up into earth orbit - and the political will to violate every nuclear and orbital weapons treaty ever to be signed!! Realistically, you'd have to construct it from lunar materials out in orbit around the moon because that's the only reasonable way to get all of that tonnage of 'stuff' up there. Orion-style spacecraft are much beloved by science fiction writers simply because they just about the only hope of reaching a significant fraction of 'c' with believable technology. SteveBaker (talk) 18:31, 7 August 2009 (UTC)[reply]

It might be better to build it out in the asteroid belt. Lots of resources out there and not even the Moon's gravity to contend with. Of course, a space elevator would make it all much easier. --Tango (talk) 19:27, 7 August 2009 (UTC)[reply]

I'll plug the book Interstellar Migration and the Human Experience, edited by Ben R. Finney & Eric M. Jones (U.Calif.Press 1985). —Tamfang (talk) 02:09, 11 August 2009 (UTC)[reply]

Birch tree with golden bark??

There is a tree groing at my workplace that I cannot identify for the life of me. It is definitvely not native to where I live and is intriguing to say the least. It cunningly resembles a birch tree and has a similar peeling type of bark. However, the bark isn't with or gray, it's golden-brown. The leafs also do not seem to be that of a birch. Anyone know what it could be? I'm going to try to attach photos later on too. It has been bugging me all summer long!! Veronika Stolbikova (talk) 15:53, 4 August 2009 (UTC)[reply]

Some eucalyptus trees shed bark, might be one of those. Is the tree indoors, or outside, and where on earth is it if it is outside? Googlemeister (talk) 16:03, 4 August 2009 (UTC)[reply]

The bark of different birch trees can vary in color. The Yellow Birch has yellow-bronze bark. The bark of the Alaska Birch ranges in color "from pure white to red, yellowish, pinkish, or gray". -- 128.104.112.100 (talk) 16:52, 4 August 2009 (UTC)[reply]

Pacific Madrone is what comes to mind for me. --jpgordon^{::==( o )} 23:33, 4 August 2009 (UTC)[reply]

why do some nuts sqeak as you eat them?

why do some nuts sqeak as you eat them? thanks. —Preceding unsigned comment added by 82.234.207.120 (talk) 16:05, 4 August 2009 (UTC)[reply]

I haven't actually noticed this myself. I imagine it is just parts of the nuts rubbing against each other, or your teeth, and the "squeak" has just something to do with the surface of the nut. --98.217.14.211 (talk) 23:12, 4 August 2009 (UTC)[reply]

It's horrible. SGGH ^ping! 11:56, 6 August 2009 (UTC)[reply]

In space

I see often in science fiction shows that people are exposed to space for maybe 5 seconds or so without spacesuits. In reality, how long would a person stay alive when exposed to space, and what would the effects be of a 5 second exposure (assuming they're "beamed up" to their spaceship almost immediately after exposure)? —Preceding unsigned comment added by 82.43.91.27 (talk) 16:05, 4 August 2009 (UTC)[reply]

See Human adaptation to space. DMacks (talk) 16:13, 4 August 2009 (UTC)[reply]

Also Space exposure. Googlemeister (talk) 16:14, 4 August 2009 (UTC)[reply]

Survival might be possible. Damage to ears or lungs is possible. Unconsciousness would occur after a few seconds. Edison (talk) 03:01, 5 August 2009 (UTC)[reply]

Not to mention decompression sickness... 98.234.126.251 (talk) 03:37, 5 August 2009 (UTC)[reply]

independent variable in x axis

hi friends,

in my basics i've learned that independent variable is marked in x axis and the dependent variable in y axis.in refrigeration i learned about T-s and P-H plots where entropy(S) and enthalpy(H) are in x axis.kindly explain me how it is made. SCI-hunter (talk) —Preceding undated comment added 16:32, 4 August 2009 (UTC).[reply]

I’m not sure I understand what the confusion is. The convention when plotting a function is to show the dependant variable vertically, and the independent variable horizontally. A function y=f(x), where y is a function of x, is plotted with y (the dependant variable) vertically, and x (the independent variable) horizontally. In a T-s plot, T is a function of s, so T (the dependant variable) is vertical, and s (the independent variable) is horizontal. In a p-h plot, p is a function of h, so p (the dependant variable) is vertical, and h (the independent variable) is horizontal. Does that help at all? Red Act (talk) 18:05, 4 August 2009 (UTC)[reply]

The above explanation is not exactly correct. Even in the simplest thermodynamic applications there will be functions of more than one variable. For instance, the ideal gas law PV=nRT can be used to obtain a relationship bewtween pressure, temperature and density for some gas, therefore one of them can be taken as a function of the other two. That means that there isn't a unique relationship between temperature and entropy that could be used to plot the temperature as a function of the entropy. More information needs to be specifyed. You could, for instance, specify what kind of thermodynamic process is being used. It could be isothermic, isobaric, isocoric, adiabatic, some other kind, or a combination of these. A carnot cycle, for instance, is given as a sequence of four different thermodynamic processes (Isothermic expansion, Adiabatic expansion, Isotermic contraction, Adiabatic contraction) which show up on a T-s plot as a rectangle, not as a function. Dauto (talk) 19:44, 4 August 2009 (UTC)[reply]

Ah. I had looked at Fig. 25.10 here, and the T-s plot and p-h plot look very much like the plots of two functions. However, reading further, I see that those two plots are actually phase diagrams, which are a different beast. The 2D phase diagrams section of that article might be a helpful read. Red Act (talk) 20:06, 4 August 2009 (UTC)[reply]

thanks a lot for the discussions on above topic.let me clear my doubt is that how could entropy become the independent variable(in case of T-S plot).we never change entropy for a system.what we usually do is to vary the temperature of a system and that causes a change in entropy of the system. SCI-hunter (talk) —Preceding undated comment added 22:34, 4 August 2009 (UTC).[reply]

No, what you do is transfer/take heat or compress/expand the system and, that way, change both the entropy and the temperature. The distinction between dependent and independent variables is often quite arbitrary since many of the functions that show up in practice can be inverted. In the T-s diagram, though, that doesn't really matter since you are not actually plotting a function. As I said, what you are doing is plotting the path taken by the system through the T-s space as some kind of thermodynamic process is being performed. Dauto (talk) 23:13, 4 August 2009 (UTC)[reply]

Keep in mind that x-independent/y-dependent convention is only that - a convention. There's no reason to have it that way, except that that's what everyone always does, and anyone looking at your graph will initially expect that's the way it's set up. As mentioned, there isn't a clear dependent/independent distinction in thermodynamics, so the first researchers set up the graphs the way that they thought looked best or made the most sense in context. (For example, the plots could have been made by a theoretician who thinks of the entropy as the core concept, and the measured temperature is simply a derived result of the intrinsic entropy of the system.) Others repeated that same sort of display in other contexts, and that then became the convention used for those graphs. Now everyone sets up T-S and P-H plots that way because that's how researchers expect the plots to be set up. By doing so their not necessarily claiming that one variable is dependent and one is independent, they're just showing the relation between the two in the conventional fashion. -- 128.104.112.100 (talk) 17:37, 5 August 2009 (UTC)[reply]

Moldy Bones

What would be the best way to wash this green, smelly stuff that I think might be a kind of mold off of some (real, animal) bones (which I think are for a comparitive collection)? I tried scrubbing with a wet toothbrush (per boss' suggestion), but that didn't really work. 138.192.58.227 (talk) 17:04, 4 August 2009 (UTC)[reply]

Like all good bosses would say "try scrubbing harder"

Alternatively why not search for "bone cleaning" - I think you need to get all the organic matter out of the bone to stop it going green/smelling.83.100.250.79 (talk) 18:13, 4 August 2009 (UTC)[reply]

You could try bleaching it with hydrogen peroxide.Calamus Fortis 18:32, 4 August 2009 (UTC)[reply]

It still needs cleaning (degreasing) first. 83.100.250.79 (talk) 19:33, 4 August 2009 (UTC)[reply]

What about boiling the bones in water? That usually removes almost everything from them. // BL \\ (talk) 22:17, 4 August 2009 (UTC)[reply]

Somewhere I read about preserving bone for bone handled knives. You have to remove all the marrow. Prolonged soaking in ammonia solution can remove the fat and protein. Caustic soda in water could do this dissolving too but is more hazardous and needs more cleanup, it will leave the mineral component behind. Graeme Bartlett (talk) 22:48, 4 August 2009 (UTC)[reply]

On prolonged boiling I would imagine caustic soda to attack the bone (it would), though it is good at defattting. Has anyone tried using washing powder - maybe a biological washing powder?83.100.250.79 (talk) 22:57, 4 August 2009 (UTC)[reply]

Sodium carbonate (washing soda) would prob'ly work pretty good. FWiW 98.234.126.251 (talk) 01:45, 5 August 2009 (UTC)[reply]

Per National Public Radio (U.S), museums use maggots to clean bones of small animals. Books on 19th century medicine say that a doctor might take human bones and put them in a wire mesh cage submerges in a pond for a year, so the small fish etc would clean them. Edison (talk) 03:00, 5 August 2009 (UTC)[reply]

A classic way to get clean bones is to use ants. At least one friend of mine has, on several occasions, taken a small dead animal and buried it in an ant hill for a couple of weeks or so. They pick the skeleton clean. This time-lapse video of ants eating a dead gecko demonstrates this ability fairly well. -- Captain Disdain (talk) 10:16, 5 August 2009 (UTC)[reply]

Most natural history museums use dermestid beetles to clean bone samples. They do an extremely thorough job, but controlling the process can be tricky simply because of how thorough they are. You likely wouldn't want them chewing your carpet away, for example. I don't know how readily ants or maggots would eat mold; I don't know if the dermestids even do that, but that's where I'd put my money. Matt Deres (talk) 16:22, 5 August 2009 (UTC)[reply]

The ants in the video not only clean the bones, but the disarticulate the skeleton and carry away some of the bones. The gecko is left in a Napoleon condition. ("bone-apart") Edison (talk) 17:18, 5 August 2009 (UTC)[reply]

electrical grounding

My son has just asked a good question (at least I thought so, with my limited scientific knowledge). I explained to him why birds don't get electrocuted while standing on an electrical wire - because the bird is not grounded and does not complete an electrical circuit. So he asked - what happens if we touch a live (exposed) wire inside an aeroplane - are we 'grounded' (earthed) or not - will we get electrocuted? Hmm! Sandman30s (talk) 19:28, 4 August 2009 (UTC)[reply]

It's all about forming a circuit, rather than grounding per se - if a (very) big bird managed to put one foot on one overhead wire, and another foot on another wire then it would get zzzapped.

The same applies in a plane or anywhere - if you make a circuit you get 'it'.

However if you touch a live wire which has the 'ground' connection also connected to the ground (ie the soil) the you can form the electrical circuit without touching two wires - the circuit being through the wire, through you, and the through the earth (soil) to the grounded connection.83.100.250.79 (talk) 19:37, 4 August 2009 (UTC)[reply]

I think the key concept here is that electrical grounding isn't "are you touching the ground?" but "are you touching something at a different voltage?" (still an approximation). Birds are OK not because they're sitting up high but because they don't provide an interesting electrical path. Even assuming an uninsulated power line, electricity will continue to flow through a short highly-conductive wire than through a long poorly-conducting bird. Since the endpoints (and end voltages) are the same, there's no reason for electricity to flow through the bird. On an airplane, though, there's a complete electrical circuit. The power supplies in the airplane are at 28 volts or 400 volts or whatever they happen to be, relative to 0 volts as defined by the plane itself. If you're touching the plane (and you are, you're in a seat or in the aisle or what have you), you're grounded, and subject to shock. You may also be interested in a past discussion of birds here or at the Straight Dope. — Lomn 19:42, 4 August 2009 (UTC)[reply]

Note also that the usual procedure for jump starting an automobile involves connecting the positive terminals of the batteries and then connecting the negative terminal of the charged battery to some metallic portion of the vehicle with the dead battery. This is called "connecting to ground" even though an automobile is actually insulated from the earth by its rubber tires. Deor (talk) 20:36, 4 August 2009 (UTC)[reply]

Your explanation of why the bird doesn't get zapped isn't correct anyway - the correct answer is actually that the bird DOES get zapped - but not by enough to bother it. The electricity has two paths it can take - the short path down an inch or so of copper wire - or the longer path: up one leg of the bird, past it's "naughty bits" and back down the other leg. The amount of current that flows down each path is inversely proportional to the resistance. Since nice thick copper wire has very little resistance - and birds have (relatively) high resistance, a huge amount of current flows down the wire - and very little of it flows through the bird. BUT that amount isn't zero! The bird is indeed being very slightly electrocuted...but the current flow, even with a soaking wet bird on a very high voltage wire is so slight that it doesn't seem to harm them. SteveBaker (talk) 22:07, 4 August 2009 (UTC)[reply]

Here's a question for you SteveBaker (without going to any reference materials!): If the line were suddenly cut so that one foot of the bird was on one side, the other foot on the other side, would the bird get electrocuted? 82.234.207.120 (talk) 23:49, 4 August 2009 (UTC)[reply]

If cutting the wire caused the two sides of the copper wire to reach different voltages, then the bird would be zapped. The crucial bit is that voltage loss along a conducting wire is "negligibly small" - so when the bird stands with two feet only inches apart on the wire, there's a small voltage drop across the bird. As Steve and others mentioned, the bird has a high resistance, so the small voltage yields an even smaller current (V = IR, or Ohm's Law). Cutting the wire could cause the two parts of the wire to reach different voltages - one half is still attached to the power source, while the other is loaded to ground (via the electric delivery network - but it might take a few seconds to decay that energy out through the grid - it's hard to say in the purely hypothetical case). But if the bird continued to stand on the two wire parts, its feet are now at very different voltages - and it will get zapped. Nimur (talk) 00:32, 5 August 2009 (UTC)[reply]

Yep. If the wire was cleanly cut - then the electricity still has two paths to choose from...through the bird - or through an inch of air. Air is a pretty good insulator - and birds are mostly salty water - which conducts electricity reasonably well (but not as well as copper wire) so the bird represents a lower resistance path than the air - so the current goes that way and the bird is undoubtedly zapped. If the process of cutting the wire allowed a spark to be produced between the ends of the wire, the air would be ionized - ionized air has a much lower resistance than regular air...so perhaps the the bird stands a better chance in that case...but I'm pretty sure the heat from the spark would still fry the poor thing. SteveBaker (talk) 01:09, 5 August 2009 (UTC)[reply]

Correct. Birds do get electrocuted by creating a complete circuit. Usually it is by touching one wing to a power source and another to ground. When I lived in the Mojave, the power poles were being redesigned because the birds there were getting zapped far too often. -- kainaw ™ 00:14, 5 August 2009 (UTC)[reply]

On power poles here, the lines are far enough apart so that birds' wings do not touch two wires. But occasionally a big bat will make that mistake, and be electocuted. Because their feet are designed to lock them upside down automatically when they are asleep, you can sometimes see electrocuted ones hanging on a wire, dead as a doornail. Bolshy birds like Currawongs will amuse themselves all day attacking this 'interloper' who should have been at home during the day. Myles325a (talk) 01:25, 5 August 2009 (UTC)[reply]

If a power line is at a sufficiently high voltage, a bird touching it even with a single foot would get a painful and perhaps dangerous jolt, even without being grounded or without touching a wire at a different voltage. The bird acts as a capacitor, and alternating current can flow through a capacitor. Higher voltage means more current. Birds will sit on power lines of 4 thousand or maybe 12 thousand volts, but I have not seen one sitting on a 69 thousand or 138 thousand volt conductor. Edison (talk) 02:57, 5 August 2009 (UTC)[reply]

Thanks for all the answers. Steve, when I google 'bird on electric wire grounding', there are lots of answers that agree with my explanation about the bird needing to be grounded, although there are just as many that talk about the potential difference/voltage between its legs. Some argue that the grounding explanation is more correct. Surely there is some compromise between these different opinions? Sandman30s (talk) 07:59, 5 August 2009 (UTC)[reply]

The potential difference between two spots an inch apart on a power line will be negligible. Take the full rated current of the conductor, multiply it by the resistance of one inch, and you have the voltage. A 4kv conductor might be bare #1/0 copper, with .105 ohms per thousand feet. One inch would be resistance of .00000875 ohm. At 300 amperes of load current, the voltage would be .0026 volts. At 1000 amps, which would overload the circuit, the voltage would still only be .009 volt across the birds feet 1 inch apart. The conductor would literally be too hot to stand on, and on its way to burning down, before it electrocuted the bird just due to the potential difference between two spots 1 inch apart. Edison (talk) 14:28, 5 August 2009 (UTC)[reply]

The "grounding" explanation is an over-simplification - and that vague and fuzzy thinking explains why you were confused about touching wires inside airplanes. Suppose there were two power lines strung parallel to each other. One is at 10,000 volts and the other one isn't quite adjusted correctly so it's at 10,100 volts. If the bird puts one foot on each wire, it'll get a 100 volt jolt and die without ever being within 50 feet of "ground". It's a matter of "potential difference" and relative resistance. If the difference between the voltages at the bird's left and right foot are sufficiently different, it dies - if not, it doesn't - irrespective of this fuzzy concept of "ground". The potential difference between 10,000 volts and "ground" is 10,000 volts - so of course the bird gets zapped if it has one foot on the ground and the other on the wire. But if you had your electricity transmission lines made out of (say) plastic instead of copper - then the bird would get zapped just standing on the plastic wire because the resistance through its body and the resistance through the plastic would be almost the same - roughly half the current would travel through the bird and it would die.

The concept of "ground" is mostly a notational matter for electrical engineers. When you talk about "grounding" a wire in a car for example - you're attaching it to the bodywork - which is typically attached to the negative side of the battery. But on some older british cars (like my '63 Mini), they opted for "Positive ground" - so the positive side of the battery is connected to the body and all of the electrical systems run on -12 volts supplied from the '-' terminal of the battery.

However, when you step out of either kind of car so that your feet are "grounded" by touching the ground - you can still get a brief but painful 'zap' (a 'static shock') from touching the door handles because the "ground" of the car isn't the same voltage as the "ground" of the planet earth. There is really no such thing as an absolute zero of voltage to use as a reference. In a sense, the bird on the 10,000 volt wire isn't getting zapped because as far as it is concerned, "ground" is the wire it's standing on.

In telecommunications and some computer applications, you have to talk about "signal ground" and "frame ground" and treat them separately because again, it's just a notational convenience. Check out (for example) RS232#Pinouts - where you'll see a "Common ground" and a "Protective ground". Note the comment there: "Use of a common ground is one weakness of RS-232: if the two devices are far enough apart or on separate power systems, the ground will degrade between them and communications will fail, which is a difficult condition to trace.". SteveBaker (talk) 14:55, 5 August 2009 (UTC)[reply]

Further... If you've worked with high-end audio equipment, you know that ground is not ground. If you assume ground is ground and connect all your equipment to the closest ground, you will get a hum on your speakers. That is because ground in one area is not necessarily the same potential as ground in another area. You need to ensure that all your equipment is using the same ground to remove difference in potential. -- kainaw ™ 15:12, 5 August 2009 (UTC)[reply]

Grounding of communication lines or control wires is indeed a confusing subject. Sometimes the ground braids are only connected to earth ground at one end of the line, to avoid ground current travelling over the braid and having a "ground loop." Edison (talk) 17:14, 5 August 2009 (UTC)[reply]

And in some power-transmission systems, the earth serves as the conductor that closes the circuit, see Single-wire earth return. For a real-world situation similar to the "one foot on each side of a cut wire"-situation mentioned above, see the end of the section Tram#Electric (trolley cars). --NorwegianBlue^talk 18:23, 5 August 2009 (UTC)[reply]

Speaking of trams... I thought that the engineer could lower the pantograph from the cab without having to jump out of the tram. Is that not so? 98.234.126.251 (talk) 04:56, 6 August 2009 (UTC)[reply]

COCONUT OIL

DOES IT CONTAIN OMEGA 3 OR 6 ? —Preceding unsigned comment added by 209.252.144.42 (talk) 21:55, 4 August 2009 (UTC)[reply]

According to the analysis at Oleic acid it contains Omega-6 Linoleic acid and Omega-9 Oleic acid. 83.100.250.79 (talk) 22:07, 4 August 2009 (UTC)[reply]

You can look it up in the USDA's nutrient database - there it says that it contains 5.8% monounsaturated fatty acids and all of this is 18:1 fatty acids (most likely all of that is oleic acid); and 1.8% polyunsaturated fatty acids and all of this is 18:2 (most likely linoleic acid) - so it seems likely that there is 1.8% omega-6 fatty acids and 0.0% omega-3 fatty acids. Icek (talk) 17:11, 9 August 2009 (UTC)[reply]

I think it's an unhealthy kind of oil. Imagine Reason (talk) 21:41, 11 August 2009 (UTC)[reply]